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CIE 9709 2023 March Paper 12 Q3

A Level / CIE / P1

CIE 9709 2023 March Paper 12 Paper · Question 3

题目

Problem

A curve has equation y=160(3x+1)2y=\dfrac{1}{60}(3x+1)^2 and a point is moving along the curve.

Find the xx-coordinate of the point on the curve at which the xx- and yy-coordinates are increasing at the same rate.

(4)
题目中文翻译

一条曲线的方程为 y=160(3x+1)2y=\dfrac{1}{60}(3x+1)^2,一个点沿曲线运动。

求曲线上 xx 坐标和 yy 坐标以相同速率增加时,该点的 xx 坐标。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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由链式法则,两个坐标以相同的正速率增加时,曲线斜率为 11。

答题过程

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Writing tt for time, dydt=(dydx)(dxdt)\frac{\mathrm{d}y}{\mathrm{d}t}=(\frac{\mathrm{d}y}{\mathrm{d}x})(\frac{\mathrm{d}x}{\mathrm{d}t}). Since the two rates are equal and positive, dydx=1\frac{\mathrm{d}y}{\mathrm{d}x}=1. Differentiating gives (3x+1)/10=1(3x+1)/10=1, so x=3x=3.